A small body slips,subject to the force of fr | vedclass

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Question

(A) The work done by friction is given by $W_f = -\int \mu N \, ds$. For route $1$,the surface is convex,so the normal force $N_1 = mg \cos 	heta - \

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speed is greater in case $1$

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(A) The work done by friction is given by $W_f = -\int \mu N \, ds$. For route $1$,the surface is convex,so the normal force $N_1 = mg \cos 	heta - \frac{mv^2}{r}$. For route $2$,the surface is concave,so the normal force $N_2 = mg \cos 	heta + \frac{mv^2}{r}$. Since $N_2 > N_1$,the frictional force $f = \mu N$ is greater on route $2$ than on route $1$. As the work done against friction is greater on route $2$,more mechanical energy is dissipated as heat on route $2$. Therefore,the final kinetic energy and speed at point $B$ will be greater for route $1$.

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